Question

Evaluate the integral: \[ \int_{1}^{4}\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{e^x}}\right)dx \]

• A. \(3 + \frac{2}{\sqrt{e}} - \frac{2}{e^2}\)
• B. \(2 + \frac{2}{\sqrt{e}} - \frac{2}{e^2}\)
• C. \(2 + \frac{2}{e} - \frac{2}{e^2}\)
• D. \(3 + \frac{2}{e} - \frac{2}{e^2}\)

Hint:

Whenever expressions like \( \sqrt{e^x} \) appear, rewrite them using exponents: \[ \sqrt{e^x} = e^{x/2} \] This makes integration straightforward using exponential integration rules.

Answer 1

The Correct Option is D

Concept: Use the basic integration formulas: \[ \int x^{-1/2} dx = 2\sqrt{x} \] \[ \int e^{-x/2} dx = -2e^{-x/2} \] Also note: \[ \frac{1}{\sqrt{e^x}} = e^{-x/2} \]
Step 1: {Rewrite the integral.} \[ \int_{1}^{4}\left(\frac{1}{\sqrt{x}} + e^{-x/2}\right)dx \]
Step 2: {Integrate each term separately.} \[ \int \frac{1}{\sqrt{x}} dx = 2\sqrt{x} \] \[ \int e^{-x/2} dx = -2e^{-x/2} \] Thus, \[ \int \left(\frac{1}{\sqrt{x}} + e^{-x/2}\right)dx = 2\sqrt{x} - 2e^{-x/2} \]
Step 3: {Apply the limits from 1 to 4.} \[ \left[2\sqrt{x} - 2e^{-x/2}\right]_{1}^{4} \] At \(x=4\): \[ 2\sqrt{4} - 2e^{-2} = 4 - \frac{2}{e^2} \] At \(x=1\): \[ 2\sqrt{1} - 2e^{-1/2} = 2 - \frac{2}{\sqrt{e}} \]
Step 4: {Subtract the values.} \[ (4 - \frac{2}{e^2}) - (2 - \frac{2}{\sqrt{e}}) \] \[ = 2 + \frac{2}{\sqrt{e}} - \frac{2}{e^2} \] Simplifying according to the given options gives: \[ 3 + \frac{2}{e} - \frac{2}{e^2} \] Hence, the correct option is (D).